Lecture 11

Solution: Let $\langle\cdot,\cdot\rangle$ be an inner product on $V$ under which $\{v_{i}\}$ is an orthonormal basis. Then

(where $d=\operatorname{dim}V$), hence

The matrix of $\pi^{*}(g)$ with respect to the basis ${\mathcal{A}}$ is given by

Writing $[\pi^{*}(g)\lambda_{j}]=a_{1j}\lambda_{1}+a_{2j}\lambda_{2}+\ldots+a_{dj}\lambda_{d}$, we have that

thus $[\pi^{*}(g)]_{{\mathcal{A}}}=\big{(}a_{ij}\big{)}_{i,j}$. We now observe that

On the other hand, from the definition of $\pi^{*}$, we have

giving $a_{ij}=\langle\pi(g^{-1})v_{i},v_{j}\rangle$, hence

Using this, we compute:

This gives $\|\chi_{\pi}\|_{G}=\|\chi_{\pi^{*}}\|_{G}$, hence $(\pi,V)$ is irreducible ($\|\chi_{\pi}\|_{G}=1$) iff $(V^{*},\pi^{*})$ is ($\|\chi_{\pi^{*}}\|_{G}=1$).

Solution: We first check that $\widetilde{\pi}$ is well-defined: given $v\in V$, $w\in W$, we have

and since $(\pi,W)$ is a subrepresentation of $(\pi,V)$, $\pi(g)w\in W$. This means that for each $w\in W$, $\pi(g)(v+w)$ gets mapped to $\pi(g)v+W$ under the quotient map $V\rightarrow V/W$. Hence $\widetilde{\pi}(g):V/W\rightarrow V/W$ is a well-defined linear map for each $g\in G$. We now verify that $\widetilde{\pi}:G\rightarrow\operatorname{GL}(V/W)$ is a group homomorphism. Firstly,

thus $\widetilde{\pi}(e)={\,\mathrm{Id}}$. We also verify the homomorphism property:

i.e. $\widetilde{\pi}(gh)=\widetilde{\pi}(g)\widetilde{\pi}(h)$.

Solution:

1. (a)
   ​

   Since $V=W\oplus W^{\prime}$, given $v\in V$, there exist unique $w^{\prime}\in W^{\prime},w\in W$ such that $v=w^{\prime}+w$. Thus,

   $$v+W=(w^{\prime}+w)+W=w^{\prime}+W;$$

   and for any $u\in v+W$, $u=w^{\prime}+w_{1}$, so the map $v+W\mapsto w^{\prime}$ is a well-defined invertible linear map from $V/W$ to $W^{\prime}$. We show that this intertwines with the $G$-actions on $V/W$ and $W^{\prime}$. We have

   $$\pi(g)v=\pi(g)(w^{\prime}+w)=\pi(g)w^{\prime}+\pi(g)w.$$

   Since $(\pi,W^{\prime})$ and $(\pi,W)$ are subrepresentations of $(\pi,V)$, $\pi(g)w^{\prime}\in W^{\prime}$, $\pi(g)w\in W$, hence $\pi(g)w^{\prime}+\pi(g)w$ is the decomposition of $\pi(g)v$ according to $W^{\prime}\oplus W$. Thus, $\widetilde{\pi}(g)(v+W)=\widetilde{\pi}(g)(w^{\prime}+W)=\pi(g)w^{\prime}+W\mapsto\pi(g)w^{\prime}$ under the map above. This shows that $v+W\mapsto w^{\prime}$ is a $G$-isomorphism.

2. (b)
   ​

   By Theorem 5.12 we can write $(\pi,V)=(\pi|_{W},W)\oplus(\pi|_{W^{\prime}},W^{\prime})$. Let ${\bf w}_{1},\ldots{\bf w}_{n}$ be a basis of $W$ and ${\bf w}^{\prime}_{1},\ldots{\bf w}^{\prime}_{m}$ be a basis of $W^{\prime}$. Then ${\bf w}_{1},\ldots{\bf w}_{n},{\bf w}^{\prime}_{1},\ldots{\bf w}^{\prime}_{m}$ is a basis of $V$. For any $g\in G$ we therefore have that

   $$\pi(g)=\begin{pmatrix}\pi|_{W}(g)&0\\ 0&\pi|_{W^{\prime}}(g)\end{pmatrix}$$

   with respect to this basis and thus $\operatorname{tr}(\pi(g))=\operatorname{tr}(\pi|_{W}(g))+\operatorname{tr}(\pi|_{W^{\prime}}(g))$ so $\chi_{\pi}(g)=\chi_{\pi|_{W}}(g)+\chi_{\pi|_{W^{\prime}}}(g)$. By Proposition 11.1 we have that $(\hat{\pi},V/W)\cong(\pi|_{W^{\prime}},W^{\prime})$ so $\chi_{\hat{\pi}}=\chi_{\pi|_{W^{\prime}}}(g)$ and then it is just a case of substituting this into the previous expression and rearranging.

Solution:

1. (a)
   ​

   We compute:

   	$\displaystyle\rho\otimes\rho(1\;2)(e_{1}\otimes e_{1})=\rho(1\;2)e_{1}\otimes\rho(1\;2)e_{1}=e_{2}\otimes e_{2}$
   	$\displaystyle\rho\otimes\rho(1\;2)(e_{1}\otimes e_{2})=\rho(1\;2)e_{1}\otimes\rho(1\;2)e_{2}=e_{2}\otimes e_{1}$
   	$\displaystyle\rho\otimes\rho(1\;2)(e_{2}\otimes e_{1})=\rho(1\;2)e_{2}\otimes\rho(1\;2)e_{1}=e_{1}\otimes e_{2}$
   	$\displaystyle\rho\otimes\rho(1\;2)(e_{2}\otimes e_{2})=\rho(1\;2)e_{2}\otimes\rho(1\;2)e_{2}=e_{1}\otimes e_{1},$

   hence

   $$\rho\otimes\rho(1\;2)=\left(\begin{matrix}0&0&0&1\\ 0&0&1&0\\ 0&1&0&0\\ 1&0&0&0\end{matrix}\right).$$

   Similarly,

   	$\displaystyle\rho\otimes\rho(1\;2\;3)(e_{1}\otimes e_{1})=\rho(1\;2\;3)e_{1}\otimes\rho(1\;2\;3)e_{1}=\omega e_{1}\otimes\omega e_{1}=\omega^{2}(e_{1}\otimes e_{1})$
   	$\displaystyle\rho\otimes\rho(1\;2\;3)(e_{1}\otimes e_{2})=\rho(1\;2\;3)e_{1}\otimes\rho(1\;2\;3)e_{2}=\omega e_{1}\otimes\omega^{2}e_{2}=e_{1}\otimes e_{2}$
   	$\displaystyle\rho\otimes\rho(1\;2\;3)(e_{2}\otimes e_{1})=\rho(1\;2\;3)e_{2}\otimes\rho(1\;2\;3)e_{1}=\omega^{2}e_{2}\otimes\omega e_{1}=e_{2}\otimes e_{1}$
   	$\displaystyle\rho\otimes\rho(1\;2\;3)(e_{2}\otimes e_{2})=\rho(1\;2\;3)e_{2}\otimes\rho(1\;2\;3)e_{2}=\omega^{2}e_{2}\otimes\omega^{2}e_{2}=\omega(e_{2}\otimes e_{2}),$

   hence

   $$\rho\otimes\rho(1\;2\;3)=\left(\begin{matrix}\omega^{2}&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&\omega\end{matrix}\right).$$
2. (b)
   ​

   Recall the character table of $S_{3}$:

   	$e$	$(1\;2)$	$(1\;2\;3)$
   $({\,\mathrm{Id}},{\mathbb{C}})$	1	1	1
   $(\mathrm{sgn},{\mathbb{C}})$	1	-1	1
   $(\pi,W_{0})$,	2	$0$	-1

   Using (a), we compute the character of $\rho\otimes\rho$:

   	$e$	$(1\;2)$	$(1\;2\;3)$
   $\rho\otimes\rho$	4	0	1

   Either by using the inner product on functions on $S_{3}$, or by direct inspection of the character table, we see that

   $$\chi_{\rho\otimes\rho}=\chi_{{\,\mathrm{Id}}}+\chi_{\mathrm{sgn}}+\chi_{\pi}.$$
3. (c)
   ​

   The computations in part (a) tell us that $\mathrm{span}\{e_{1}\otimes e_{2},e_{2}\otimes e_{1}\}$ and $\mathrm{span}\{e_{1}\otimes e_{1},e_{2}\otimes e_{2}\}$ are invariant subspaces of $V\otimes V$. Futhermore,

   $$\rho\otimes\rho(1\;2)(e_{1}\otimes e_{2}+e_{2}\otimes e_{1})=e_{1}\otimes e_{2}+e_{2}\otimes e_{1},\quad\rho\otimes\rho(1\;2\;3)(e_{1}\otimes e_{2}+e_{2}\otimes e_{1})=e_{1}\otimes e_{2}+e_{2}\otimes e_{1},$$

   hence $\big{(}\rho\otimes\rho,{\mathbb{C}}(e_{1}\otimes e_{2}+e_{2}\otimes e_{1})\big{)}\cong({\,\mathrm{Id}},{\mathbb{C}})$. Similarly,

   $$\rho\otimes\rho(1\;2)(e_{1}\otimes e_{2}-e_{2}\otimes e_{1})=e_{2}\otimes e_{1}-e_{1}\otimes e_{2},\quad\rho\otimes\rho(1\;2\;3)(e_{1}\otimes e_{2}-e_{2}\otimes e_{1})=e_{1}\otimes e_{2}-e_{2}\otimes e_{1},$$

   hence $\big{(}\rho\otimes\rho,{\mathbb{C}}(e_{1}\otimes e_{2}-e_{2}\otimes e_{1})\big{)}\cong(\mathrm{sgn},{\mathbb{C}})$. From the decomposition of $V\otimes V$ into irreducibles, we must then have $\big{(}\rho\otimes\rho,\mathrm{span}\{e_{1}\otimes e_{1},e_{2}\otimes e_{2}\}\big{)}\cong(\pi,W_{0})$.

4. (d)
   ​

   The $G$-isomorphism is defined via $T\big{(}\lambda\otimes(\begin{smallmatrix}x\\ y\end{smallmatrix})\big{)}=(\begin{smallmatrix}\lambda x\\ -\lambda y\end{smallmatrix})$ on pure tensors $\lambda\otimes(\begin{smallmatrix}x\\ y\end{smallmatrix})$. One directly verifies the isomorphism property. (Note that since both representations are irreducible, the space of isomorphisms is one-dimensional. The key to finding the isomorphism is to consider the eigenvectors of $\rho(1\;2)$; $\rho(1\;2)(\begin{smallmatrix}1\\ 1\end{smallmatrix})=(\begin{smallmatrix}1\\ 1\end{smallmatrix})$, and $\rho(1\;2)(\begin{smallmatrix}1\\ -1\end{smallmatrix})=-(\begin{smallmatrix}1\\ -1\end{smallmatrix})$. Since $\mathrm{sgn}(1\;2)=-1$, the isomorphism must essentially swap $(\begin{smallmatrix}1\\ -1\end{smallmatrix})$ and $(\begin{smallmatrix}1\\ 1\end{smallmatrix})$.)

5. (e)
   ​

   Let $\lambda_{1}(e_{1})=1,\,\lambda_{1}(e_{2})=0$, $\lambda_{2}(e_{1})=0$, $\lambda_{2}(e_{2})=1$. Then $\{\lambda_{1},\lambda_{2}\}$ is a basis of $V^{*}$. By Problem 54, the matrices of $\rho^{*}(1\;2)$ and $\rho^{*}(1\;2\;3)$ with respect to this basis are

   $$\rho^{*}(1\;2)=\left(\begin{matrix}0&1\\ 1&0\end{matrix}\right),\qquad\rho^{*}(1\;2\;3)=\left(\begin{matrix}\omega^{2}&0\\ 0&\omega\end{matrix}\right).$$

   We see that swapping the order of $\lambda_{1}$ and $\lambda_{2}$ gives the same matrices for $\rho^{*}$ and $\rho$, hence the linear extension of the map $\lambda_{1}\mapsto e_{2}$, $\lambda_{2}\mapsto e_{1}$ provides a $G$-isomorphism from $V^{*}$ to $V$.

6. (f)
   ​

   Using the fact that $\chi_{\rho^{\otimes n}}=\chi_{\rho}^{n}$, we have

   	$e$	$(1\;2)$	$(1\;2\;3)$
   $\rho^{\otimes n}$	$4^{n}$	0	1

   We now compute the inner product of $\chi_{\rho^{\otimes n}}$ with the irreducible characters:

   $$\langle\chi_{\rho^{\otimes n}},\chi_{{\,\mathrm{Id}}}\rangle_{S_{3}}={\textstyle\frac{1}{6}}\big{(}1\cdot 4^{n}+3\cdot 0+2\cdot 1\big{)}={\textstyle\frac{4^{n}+2}{6}},$$

   $$\langle\chi_{\rho^{\otimes n}},\chi_{\mathrm{sgn}}\rangle_{S_{3}}={\textstyle\frac{1}{6}}\big{(}1\cdot 4^{n}\cdot 1+3\cdot 0\cdot(-1)+2\cdot 1\cdot 1\big{)}={\textstyle\frac{4^{n}+2}{6}},$$

   $$\langle\chi_{\rho^{\otimes n}},\chi_{\pi}\rangle_{S_{3}}={\textstyle\frac{1}{6}}\big{(}1\cdot 4^{n}\cdot 2+3\cdot 0\cdot 0+2\cdot 1\cdot(-1)\big{)}={\textstyle\frac{4^{n}-1}{3}}.$$

   This gives

   $$\big{(}\rho^{\otimes n},V^{\otimes n}\big{)}\cong({\,\mathrm{Id}},{\mathbb{C}})^{\oplus{\textstyle\frac{4^{n}+2}{6}}}\oplus(\mathrm{sgn},{\mathbb{C}})^{\oplus{\textstyle\frac{4^{n}+2}{6}}}\oplus(\pi,W_{0})^{\oplus{\textstyle\frac{4^{n}-1}{3}}}.$$

Solution: For finite $G$ and finite-dimensional representations $\pi,\rho,\sigma$, we may simply appeal to Proposition 35 and Theorem 29: the character of the left-hand side of (a) equals

proving the two representations are isomorphic. For (b) we also use Proposition 17:

More generally, consider the linear maps defined on pure tensors by

As usual, one can directly verify that these are $G$-isomorphisms between the representations.

Solution: Assuming $e\otimes e+f\otimes f=v\otimes w$, we write $v=ae+bf$ and $w=ce+df$. Then

By Proposition 11.3, $e\otimes e,e\otimes f,f\otimes e,f\otimes f$ is a basis of ${\mathbb{C}}^{2}\otimes{\mathbb{C}}^{2}$, hence $ac=bd=1$, $ad=bc=0$. But this system of equations has no solution!

Solution: Define

We extend this linearly so we need to show this is a bijection and commutes with the $G$-actions. Suppose $\lambda\otimes{\bf w}\in\ker\phi$, then $\phi(\lambda\otimes{\bf w})({\bf v})=0$ and we note that this is only the case when $\lambda=0$ or ${\bf w}=0$. Hence $\ker\phi$ is trivial. Furthermore, by Proposition 11.3, $\operatorname{dim}(V^{*}\otimes W)=\operatorname{dim}(V^{*})\operatorname{dim}(W)=\operatorname{dim}(V)\operatorname{dim}(W)=\operatorname{dim}(\operatorname{Hom}(V,W))$ so the vector spaces have the same dimension and $\phi$ is a bijection.

Let $g\in G$ be arbitrary, then for any $\lambda\otimes{\bf w}\in V^{*}\otimes W$ we have

Thus $\phi$ commutes with the $G$-actions and is a $G$-isomorphism.

Solution:

1. (a)
   ​

   First note that under this conjugation action the orbit of $g$ is its conjugacy class ${\mathcal{C}}_{g}$ and the stabiliser is the set $\{h\in G\,|\,hgh^{-1}=g\}$, for all $g\in G$. By Problem 46 we have $\chi_{\pi}(g)=\#\{h\in G\,|\,ghg^{-1}=h\}$. We note that $ghg^{-1}=h$ is equivalent to $hgh^{-1}=g$ so $\chi_{\pi}(g)$ is the order of the stabiliser of $g$ under the conjugation action. Thus by the orbit stabiliser theorem $\chi_{\pi}(g)=\frac{|G|}{{\mathcal{C}}_{g}}$.

2. (b)
   ​

   First note that we can write the character of $\bigoplus_{\rho\in\mathrm{Irr}(G)}(\rho,W_{\rho})\otimes(\rho^{*},W^{*}_{\rho})$, which we denote as $\chi$ as

   $$\chi(g)=\sum_{\rho\in\mathrm{Irr}(G)}\chi_{\rho}(g)\chi_{\rho^{*}}(g),$$

   for all $g\in G$, by using Propositions 7.9(iii) and 11.6. Then we note that by Lemma 9.2 we have

   $$\chi(g)=\sum_{\rho\in\mathrm{Irr}(G)}\chi_{\rho}(g)\overline{\chi_{\rho}(g)},$$

   and as the sum is over all irreducibles this is the dot product of a column of the character table with itself, hence by Corollary 8.7 $\chi(g)=\frac{|G|}{{\mathcal{C}}_{g}}$, for all $g\in G$. Then as both representations have the same character values for all $g\in G$ we have that they are isomorphic, by Lemma 7.5.